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4hv.org :: Forums :: General Science and Electronics
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Modified Sine Wave Inverter
Move Thread LAN_403
Avalanche
Sat Apr 01 2006, 07:19AM Print
Avalanche Registered Member #103 Joined: Thu Feb 09 2006, 08:16PM
Location: Derby, UK
Posts: 845
 		Just wanted to know this for a future project.

When you see a modified sine wave inverter, I know it has a waveform like the following, but what about duty cycle? Is it really supposed to be 50% as depicted in the picture (attached)?

I want to make a 'cheap and nasty but high power' inverter, using a fullbridge of IRFP460s on the output stage. It's a future project, but should be good for powering things like lights, and appliances that contain switch mode power supplies.

Thanks
1143875990 103 FT0 Modsine
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Wilson
Sat Apr 01 2006, 08:29AM
Wilson Registered Member #78 Joined: Thu Feb 09 2006, 11:27AM
Location: Sydney, Australia
Posts: 133
 		The waveform is adjusted so the rms value matches that of a standard sine wave.. I don't see any reason why it wouldn't be a 50:50 DC.
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Arjan EMM
Sat Apr 01 2006, 10:55AM
Arjan EMM Registered Member #149 Joined: Sun Feb 12 2006, 09:11AM
Location: Netherlands
Posts: 12
 		50% will give the same peak and rms voltage as the sinewave

2          2     
    /~  (0.5*0  + 0.5*160  )    
\  /    (------------------)= 113 Vac RMS
 \/     (    0.5 + 0.5     )
Just as a 113Vac rms sine gives 113*1,41=160 peak
Link2

There is a little error in your drawing btw. The peak of the sine should be the same height as the top of the square wave.

Edit: For some strange reason the math bit keeps changing appearance. I can't find anything here on the site about using preformatted text so i can't fix it. Checking back a few minuts later seems to help :P
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Avalanche
Sat Apr 01 2006, 11:22AM
Avalanche Registered Member #103 Joined: Thu Feb 09 2006, 08:16PM
Location: Derby, UK
Posts: 845
 		Thanks for the replies!

I will no doubt post this on the project board when I get around to building it
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